Does a “composite field” always exist?

Suppose $F$ is a field, and $F_1, F_2$ are two extension fields of $F$. Is it always the case that there is a field $L$, containing three subfields $F, K_1, K_2$ and two ring isomorphisms $\varphi_{i}:F_i\rightarrow K_1$ fixing $F$?

Note 1: We lose no generality assuming $F$, rather than an isomorphic copy of $F$, is a subfield of $L$.

I ask this because I was wondering if there is a way to combine the reals and the $p$-adic numbers into a single extension of $\mathbb{Q}$.

Note 2: I seem to recall someone telling me this couldn't be done (perhaps with additional topological data preserved). But I cannot seem to remember the reason why. In any case, I want to know if there is something other than topology which prevents it.

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Concerning the p-adics, reals: We have $\mathbb{R} \subseteq \mathbb{C}$, $\mathbb{Q}_p \subseteq \mathbb{C}_p$ and $\mathbb{C} \cong \mathbb{C}_p$. Using this isomorphism you can embedd $\mathbb{Q}_p$ into $\mathbb{C}$. Of course this iso. isn't defined in a constructible way. – Ralph Feb 6 '12 at 20:12
Regarding Note 2: There is no topological field containing topological copies of $\mathbb{R}$ and $\mathbb{Q}_p$, since each of these induce distinct topologies on $\mathbb{Q}$. The isomorphism Ralph describes is not continuous. – Kevin Ventullo Feb 7 '12 at 3:05

The tensor product $F_1 \otimes_F F_2$ is not 0, hence it has a quotient which is a field. This contains the images of both $F_i$.

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Years ago, I first learnt the solution in David's answer and could not really find anything enlightening or memorable about it. Later the argument using tensor products was used in a text on algebraic geometry (namely the lemma that $|X \times_S Y| \to |X| \times_{|S|} |Y|$ is surjective) and of course now everything was clear as crystal. As a side remark, both proofs use variants of the axiom of choice (even twice). – Martin Brandenburg Feb 7 '12 at 11:00
(This also reminds me of the concise tensor product construction of the algebraic closure of a field: In $k' := \bigotimes_{0 \neq f \in k[x]} k[x]/(f)$ every polynomial as a root, thus the colimit of $k \subseteq k' \subseteq k'' \subseteq k''' \subseteq ...$ is an algebraic closure of $k$. One can show $k''=k'$, but this is not trivial.) – Martin Brandenburg Feb 7 '12 at 11:03

Sure. Find $T_i$ between $F$ and $F_i$ such that $T_i/F$ is pure transcendental and $F_i/T_i$ is purely algebraic. Let $T_i = F(S_i)$, with the $S_i$ algebraically independent. Without loss of generality, suppose that the cardinality of $S_1$ is less than or equal to that of $S_2$. Then the algebraic closure of $F(S_2)$ is a suitable $L$.

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In the language of Model Theory, your question can be rewritten as: "does the theory of fields have the amalgamation property? and the answer is yes.
Well known examples of theories with the amalgamation property include: fields, ordered fields, groups, abelian groups and boolean algebras.

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+1 since you provide the general background, but it would be even a better answer if you add a specific reference where the amalgamation property for fields is proved. I believe that model theorists argue as in David's answer or equivalently with "$\kappa$-categorical" arguments. – Martin Brandenburg Feb 7 '12 at 11:07
@Martin: To be honest, I´ve never seen a published proof of the fact that fields have the amalgamation property. Model Theory textbooks (e.g. Hodges, Chang & Keisler) just mention the fact and use it to prove other things, like quantifier elimination for algebraically closed fields. – Ramiro de la Vega Feb 7 '12 at 11:53